Coulomb Application

1) depends on if the thread is uniformly charged. (Which will not always be the case)

If the thread is not charged uniformly, it will not take a circular shape..

Otherwise there will be "symmetric" repulsions that will cause it to become circular.

The second part is the more interesting one..

you have to take an element of angle d theta from the center.

The charge on that element will be Q/(2pi R) . (R dθ) = Q (dθ/2pi)

Now draw the fbd of the element.

The force on the element will be
Initilaly if force F0 acts on the elemtn, then
F₀=2(T) sin(dθ/2)
Later force acting increases due to the charge at the center
k Q (dθ/2pi)Q₀r²+F₀=2(T+ΔT) sin(dθ/2)

SInce dθ is very small, you can replace sin by the angle itself.

so

k Q (dθ/2pi)Q₀r²=2ΔT sin(dθ/2)

k Q Q₀2pi r²=ΔT

The image of element

THe trick here is that i never found F₀

If you see closely, it got cancelled out with the 2nd equation..

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